Problem: The equation of an ellipse $E$ is $\dfrac {(x-6)^{2}}{4}+\dfrac {(y+4)^{2}}{49} = 1$. What are its center $(h, k)$ and its major and minor radius?
Answer: The equation of an ellipse with center $(h, k)$ is $ \dfrac{(x - h)^2}{a^2} + \dfrac{(y - k)^2}{b^2} = 1$ We can rewrite the given equation as $\dfrac{(x - 6)^2}{4} + \dfrac{(y - (-4))^2}{49} = 1 $ Thus, the center $(h, k) = (6, -4)$ $49$ is bigger than $4$ so the major radius is $\sqrt{49} = 7$ and the minor radius is $\sqrt{4} = 2$.